1991 OIM Problems/Problem 4

Problem

Find a number $N$ of five different and non-zero digits, which is equal to the sum of all the numbers of three different digits that can be formed with five digits of $N$ .

~translated into English by Tomas Diaz. ~orders@tomasdiaz.com

Solution

Let $N=d_1d_2d_3d_4d_5$ or in a better format: $N=10000d_1+1000d_2+100d_3+10d_4+d_5$

The total number of combinations is given the following way:

For the first digit of any three-digit number we have 5 numbers to chose from.

For the second digit we have 4 numbers to chose from.

For the third digit we have 3 numbers to chose from.

Total numbers of three digit numbers is (5)(4)(3)=60.

Now we need to find their sum.

From all 60 ways, in the first digit we will have each digit of N showing with (4)(3)=12 configurations.

From all 60 ways, in the second digit we will have each digit of N showing with (4)(3)=12 configurations.

From all 60 ways, in the last digit we will have each digit of N showing with (4)(3)=12 configurations.

Therefore the sum, since each digit of $N$ is shown in each position 12 times, then

$S=\left( 12\sum_{i=1}^{5}d_i \right)100+\left( 12\sum_{i=1}^{5}d_i \right)10+\left( 12\sum_{i=1}^{5}d_i \right)$

$S=1332 \sum_{i=1}^{5}d_i =N$

Since $12345\le N \le 98765$ , then $15 \le \sum_{i=1}^{5}d_i \le 35$

$\begin{cases}(15)(1332)=19980; 1+9+9+8+0=27; 27 \ne 15; \text{NO} \end{cases}$

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

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1991 OIM Problems/Problem 4

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